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Unformatted text preview: Initially, Q = 4 = 2 E = = dq = 8 / 3 E total = 8/3 When t > , Q = Q 1 + Q 2 2 = + (1 x V) V = 1 = dq + 0.5 x C x V 2 E total, = 1/3 + 5/2 = 5/6 E discharged = E total , 0 E total, = 8/3 5/6 = 1.83 J 3) Find the amplitudes of the sinusoids at Vo1 and Vo2 when vi is applied The emitter current I E is fixed at 1 mA. So whatever be the change in the base voltage, an ideal current source keeps the current in its path constant. Hence the small signal change in emitter current i e = 0 Therefore the small signal base current i b = i c / ( + 1) = 0. Hence v Hence v be = 0 V. Here v b = v i and v e = v o2 . Therefore v o2 = v i. Also collector current i c = x i e = 0 . Hence v o1 =  i c x R = 0V...
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 Fall '10
 ABIDI

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